# Abstract Algebra – Subgroup of Finite Group of Order 936

abstract-algebrafinite-groupsgroup-theorysylow-theory

I'm very confuse with this exercise. The prime number decomposition of $$|G|$$ is $$936=2^{3}3^{2}13$$. That is, $$G$$ is not a $$p$$-group, has no order $$pq$$, $$p^2q$$, $$p^mq$$ or $$p^mq^n$$. And $$|H|=117=3^{2}13$$ is not a prime or power of prime. And $$G$$ is not a cyclic group. In other words, it seems to me that $$G$$ and $$H$$ do not fit any of the conditions and consequences of Sylow's theorems and their corollaries. Any tips on what to do? Is there a theorem that fits these conditions that I have missed?

By the Sylow Theorems, the number of $$13$$-Sylow subgroups must divide $$|G|$$ and be congruent to $$1$$ modulo $$13$$. So it is a divisor of $$2^3\times 3^2$$ that is congruent to $$1$$ modulo $$13$$.

You have possible values $$1$$, $$2$$, $$4$$, $$8$$, $$3$$, $$6$$, $$12$$, $$24$$, $$9$$, $$18$$, $$36$$, and $$72$$. Of these, the only one that is congruent to $$1$$ modulo $$13$$ is $$1$$. So $$G$$ has a unique, hence normal, subgroup of order $$13$$, $$P$$.

Since $$P$$ is normal, for every subgroup $$K$$ of $$G$$, we have $$KP=PK$$, hence $$KP$$ is always a subgroup of $$G$$. The order is $$|KP| = |K||P|/|K\cap P|$$. Since $$|P|=13$$, all you need is a subgroup of order $$9$$. Does $$G$$ have a subgroup of order $$9$$?